Begin of SLS-2-2. How RC cracks 10/04/2018

Transcription

1 Begin of SLS-2-2 How RC cracks Firstly, it is important to understand the process of RC cracking. For the simplicity, it is explained on RC member in pure tension. For bending, it is in principle the same. There are three important phases, there: Phase I uncracked - fig a) both concrete and reinforcement act fully in tension and there are no cracks in concrete. Of course, acting force F 1 is relatively small. (s c <f ctk ) Phase II cracked (partly) - fig b) first cracks occur and with increasing of F 2 occurs new an new cracks until the situation on fig. c) is reached. From this point, no new crack occurs and with increasing F 3 grows the strain of reinforcement in cracks, only. It is phase III. - inelastic (fully cracked)(f 1 < F 2 < F 3 ) 1

2 More details of stress strain conditions surrounding a crack in the phase 2. or 3. le S 0 anchoring length Demonstration of previous cracking sequence in L-D diagram 2

3 Transformation of L-D diagram into L-D common relation according to Eurocode 2 NOT ESSENTIAL Influence of acting bending moment M on stiffness B of RC cross-section s r 3

4 By real structures the most stressed CS are usually in phase II so usage of interpolation is actual. The question is, which quantity to interpolate for expression the actual stiffness of related CS? Usually By curvature the real (1/r) service or load directly of RC the structures/members stiffness B is used. phase Interpolation ll is presumed formula equivalent for stiffness B is: must be interpolated. The formula is: 1 = [2] B B I B II Where: B I = E cm. J I and B Il = E cm. J Il = 1. ( s sr ) 2 s s interpol. coefficient (for tensional fastening) s sr is stress in the reinforcement after the first crack occurs s s is stress in the reinforcement for load considered. Stress can be substituted with M: = 1. ( M cr ) 2 M E M E acting M of loading (loading case) considered is coefficient for load duration = 1,0 for short term load = 0,5 for long term variable or permanent load M E > M cr by phase II! M cr (cracking moment) should be calculated for the ideal/transformed CS without crack! Basic theory of elasticity should be used and stress in marginal fibres in tension is presumed to be equal to f ct,k; 0,05. Deflections and proper loading combination Finding of relevant load combination for deflection calculation is little bit complicated. As the cracking should be presumed as irreversible process, for calculation of M cr and all stress values concerning stiffness B, B l, B ll characteristic load combination should be used (equation 6.14b by EN 1990). For calculation of the deflection itself, frequent or quasipermanent combination should be used, in relation to the type of calculated deflection. For short-term deflection (without influence of creep and shrinkage) frequent combination should be used. For long-term deflection (with influence of creep and shrinkage) quasi-permanent combination should be used. 4

5 Long-term deflection and long-term effects The long-term deflections are significantly greater when compared to short-term ones (about 2-3-times). The responsible processes of the increase are creep of concrete and shrinkage. Creep influence Is expressed simply by modification of the concrete s modulus of elasticity. Instead of E cm is E cm,eff used. Hold true: E cm,eff = E cm /(1+,t0 ) The value of E cm,eff with help of a nomogram in the next page figure can be found. e,eff = E s /E cm,eff The unpleasant thing is that all calculation must be done twice, of stiffness B l and B ll and their interpolation with the E cm,eff and e,eff. It is not enough to put E cm,eff it into deflection calculation/formula, only. Example: Calculate by the slab with h=300 mm (dried from both surfaces), concrete C30/37 with CEM N (normal setting speed) the value of,t0, the relative humidity of surrounding air is 50%. 1. step: to choose proper diagram ( there are more diagrams in EN, there] In the diagram it is h 0 = 2A C /u, where: A C is the sectional area of the CS and u the perimeter exposed to drying. The value of,t0 = 2,85 5

6 Shrinkage influence to deflection (curvature) Concrete shrinkage has the influence to deflection of members due to not uniform drying or/and not symmetrical reinforcement. Examples: In the case of symmetrical CS and also reinforcement is shrinkage symmetrical to centroid axis of the member, which shrinks but there is no curvature. In the case of not symmetrical CS and/or reinforcement shrinkage o concrete lead to curvature deflection of the member. Pay attention to the orientation of the curvature! The deflection due to external load can be increased or decreased! Shrinkage formulas Curvature due to shrinkage: 1/r cs = cs. e,eff. S/J Where: cs is the magnitude of free concrete shrinkage in actual humidity temperature and time.(formulas in the EN ) S J = S I. 1 + S II. J I J II Deflection due to shrinkage can be obtained as value of bending moment M of the same CS of member loaded with the relevant curvature. E.g. for s. s. beam the max deflection in the middle is: f sh = l 2 r sh Note: deflection due to shrinkage should be about one order smaller when compared to the creep deflection. 6

7 Stiffness of the whole members In the previous text the calculation of stiffness in relation to the level of stress in actual CS. By members stressed with bending moment M is the value of stiffness valid just in this CS. By other CS wit lover M is the stiffness equally higher see enclosed figure and M-B relation. Full respect to this fact will complicate the calculation of deflection a lot. Fortunately is by the EN allowed safe simplification to presume by the member alongside the whole span constant stiffness equal to the minimum found in the most stressed CS. Task: analyse yourself from the same point of view shape of stiffness line alongside of primary beam of frame structure. Deflection control due to comparison with max span/depth ratio This technique of indirect deflection control is in the Eurocode recommended over the previous one. The technique is based on calculation of member s span/depth ratio and comparing them with recommended limit. The calculation of an actual ratio l/d is not complicated. More complication can be presumed by usage of complicated formulas for calculation of the limit (max allowed) l/d see the following slide do not memorize!! This is the reason why by various code companion literature instead of these graphs have been presented. See over next slide. 7

8 max Do not memorize formulas and following graph!! To be honest, the values of limiting span/depth ratio must be multiplied with the values of k given in Table 7.4N in Eurocode. 8

9 Practical loading history Identified quantities without creep influence 9

10 END OF SLS Possible not obligatory plus for exam 10